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On Apr 10, 7:52*am, Cecil Moore wrote:
Keith Dysart wrote:
But the distributed capacitance and inductance
are physical impedances.

But they are constant, i.e. there is no physical
impedance *discontinuity*. The reflection coefficient
inside a homogeneous piece of transmission line is
(Z0-Z0)/(Z0+Z0)=0, i.e. there can be no reflections.
The reflection coefficient in free space is
(1.0-1.0)/(1.0+1.0)=0, i.e. there can be no reflections
in free space.

Neither 'virtual impedance' nor 'impedance, virtual'
are in the dictionary (at least the 7th Edition).

"Virtual" essentially means that no physical impedor
exists. The virtual impedance definition is covered
by definition (B), the ratio of voltage to current
which *causes* the impedance. A virtual impedance
is an *effect*, not a cause.

The transmission line definitely falls into
definition (C), "A physical device or combination of
devices whose impedance as defined in definition (A) or
(B) can be determined." The TL is a combination of
devices, a lot of very small ones, and its impedance
can be determined.

Using 26 pf/ft as a representative value for
RG-58, dividing the 45 degree section into 45
pieces, applying the normal rules for parallel and
series circuit elements, the impedance at the
entry to the line is trivially (using Excel) calculated
to be 50.443 /_ 90. Subdividing into smaller
elements would increase accuracy. If I could remember
my calculus, the exact answer could be derived.

There is no need for forward or reflected waves
at all; just basic AC circuit theory.

...Keith

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On Apr 10, 8:01*am, Cecil Moore wrote:
Keith Dysart wrote:
Finer grained analysis shows that the imputed
energy (not average) in the reflected wave is not
dissipated in the source resistor.

It is the joules in instantaneous power that must
be conserved, not the instantaneous power. There
is no such thing as a conservation of power
principle yet all you have presented are power
calculations. "Where's the beef?"

The computation using energy instead of power has
also been done (and published here) and found also
to demonstrate that the reflected is not dissipated
in the source resistor.

How many joules are there in 100 watts of
instantaneous power?

Obviously. It depends on how long you let the
100 W of instantaneous power flow. Integrate and
the answer shall be yours.

...Keith

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On Apr 10, 9:01*am, Cecil Moore wrote:
Keith Dysart wrote:
As I expected, you claim that the situations are
*completely* different. And yet the voltage, current
and energy distributions are identical. There are
no observable differences. And yet you claim they
are *completely* different. And yet there are no
observable differences. And yet....

I did NOT claim that the situations are *completely*
different. I said that some conditions are different
and some conditions are the same. Voltages and currents
are the same yet there is certainly a difference between
an open circuit and a short circuit. Besides, in the
real world, cutting the line would certainly cause
observable differences.

Tis a puzzle, isn't it.

Nope, if you were born without your five senses,
you would feel that way about everything in existence.
Why do you deliberately choose to remain handicapped
by ignorance?

A bit of modulation would cure up the mystery for
you. If any modulation crosses the node, it is a
good bet that wave energy is carrying the
modulation.

When you modulate the carrier, the resulting signal
has many frequencies. Unless great care is taken
with the choice of modulation frequencies and
carrier frequencies, they will each have nodes
at different places on the transmission line. Without
a common node, there is no place that energy
does not cross, hence modulation makes the
question moot. In effect, the standing waves
of the various frequency components do not line
up. (Remember superposition works, for voltages).

If care is taken with the selection of modulation
frequencies with regards to the carrier, then nodes
can be created on the transmission line and neither
the carrier nor the modulation will cross such a
node.

If phase locked TV signal generators
equipped with circulator load resistors are
installed at each end of a transmission line,
the TV signals can be observed on normal TV
sets crossing the standing wave nodes as if
they didn't exist.

Which they don't, as explained above.

Removing the modulation
is unlikely to reverse the laws of physics.

True, but it can change whether there are nodes.

And we know from circuit theory that we can cut
a conductor carrying no current without affecting
the circuit. Why should it be different here?

Please prove that a short circuit and an open circuit
are identical.

An open circuit is just as useful as a short circuit
when no current is flowing.

Please present your new laws of physics that allow
EM waves to reflect off of EM waves in the complete
absence of a physical discontinuity.

Again, not my claim.

Seems your theory requires such. Please explain how
reflections can occur at a passive standing wave node
without EM waves bouncing off of each other.

Energy and momentum both must be conserved. A causeless
reversal of energy and momentum is impossible whether
it is a bullet or an EM wave.

What causes energy to flow into and then out of a capacitor?
Look for your answer there.

But using your previous approach
for analysis, perhaps we should insert a zero length
line of the appropriate impedance to provide the cause
for the reflection, if you insist on a reflection.

Please produce an example of
a real world transmission line that would support your
100% reflection. Hint: what would be the Z02
characteristic impedance in the reflection
coefficient equation, (50-Z02)/(50+Z02) = 1.0 ???

That is just too easy...
(50-Z02)/(50+Z02) = 1.0
50 - Z02 = 50 + Z02
-2 * Z02 = 0
Z02 = 0
That wasn't so hard, was it?

...Keith

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Keith Dysart wrote:
There is no need for forward or reflected waves
at all; just basic AC circuit theory.

Now do it in free space. EM waves are EM waves
no matter what the medium.
--
73, Cecil http://www.w5dxp.com

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Keith Dysart wrote:
The computation using energy instead of power has
also been done (and published here) and found also
to demonstrate that the reflected is not dissipated
in the source resistor.

Well, that certainly violates the conservation of
energy principle. We know the reflected energy is
not dissipated in the load resistor, by definition.

The only other device in the entire system capable
of dissipation is the source resistor. Since the reflected
energy is not dissipated in the load resistor and you say
it is not dissipated in the source resistor, it would
necessarily have to magically escape the system or build
up to infinity (but it doesn't). You keep digging your
hole deeper and deeper.

How many joules are there in 100 watts of
instantaneous power?

Obviously. It depends on how long you let the
100 W of instantaneous power flow. Integrate and
the answer shall be yours.

I'm not the one making the assertions. How many joules
of energy exist in *YOUR* instantaneous power calculations?
--
73, Cecil http://www.w5dxp.com

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Keith Dysart wrote:
If care is taken with the selection of modulation
frequencies with regards to the carrier, then nodes
can be created on the transmission line and neither
the carrier nor the modulation will cross such a
node.

Please prove your assertion on the bench. Until you
do, there is little left to discuss.

Please produce an example of
a real world transmission line that would support your
100% reflection. Hint: what would be the Z02
characteristic impedance in the reflection
coefficient equation, (50-Z02)/(50+Z02) = 1.0 ???

That is just too easy...
(50-Z02)/(50+Z02) = 1.0
50 - Z02 = 50 + Z02
-2 * Z02 = 0
Z02 = 0
That wasn't so hard, was it?

Now build one. Be sure to verify that you can transfer
energy from end to end. Until you do, there is little
left to discuss.
--
73, Cecil http://www.w5dxp.com

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Cecil Moore wrote:
Keith Dysart wrote:
If care is taken with the selection of modulation
frequencies with regards to the carrier, then nodes
can be created on the transmission line and neither
the carrier nor the modulation will cross such a
node.

Please prove your assertion on the bench. Until you
do, there is little left to discuss.

And if you do, the distributed network model will
have to be overhauled.
--
73, Cecil http://www.w5dxp.com

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On Fri, 11 Apr 2008 13:25:40 GMT
Cecil Moore wrote:

Keith Dysart wrote:
The computation using energy instead of power has
also been done (and published here) and found also
to demonstrate that the reflected is not dissipated
in the source resistor.

Well, that certainly violates the conservation of
energy principle. We know the reflected energy is
not dissipated in the load resistor, by definition.

The only other device in the entire system capable
of dissipation is the source resistor. Since the reflected
energy is not dissipated in the load resistor and you say
it is not dissipated in the source resistor, it would
necessarily have to magically escape the system or build
up to infinity (but it doesn't). You keep digging your
hole deeper and deeper.


You write "The only other device in the entire system capable
of dissipation is the source resistor." which is a correct statement. Unfortunately, the circuit is intended to illustrate the absence of interference under special circumstances but an instant analysis shows that all the power can not be accounted for. We can only conclude that interference is present. Not good because the circuit was intended to illustrate a case of NO interference.

Our choice of a voltage source is incomplete because we did not assign it a mechanism to provide a reactive voltage, allowing the source to only apply a sinsoidal voltage without specifying the current or current timing. As a result, reflected power will return to the source resulting in an apparent loss of power to the system and resistor Rs. It is not a magical loss of power, only the result of interference acting within the cycle.

The circuit is very useful to investigate interference more carefully because on the AVERAGE, the interference IS zero. Using spreadsheets, we can see how the interference both adds and subtracts from the instantaneous applied voltage, resulting in cycling variations in the power applied to the resistor and other circuit elements. A very instructive exercise.

--
73, Roger, W7WKB

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Roger Sparks wrote:
You write "The only other device in the entire system capable
of dissipation is the source resistor." which is a correct statement.

Therefore, all power dissipated in the circuit must be dissipated
in the load resistor and the source resistor because there is
nowhere else for it to go. Since the reflected power is not
dissipated in the load, by definition, it has to be dissipated
in the source resistor but not at the exact time of its arrival.
There is nothing wrong with delaying power dissipation for 90
degrees of the cycle. In Parts 2 and 3 of my articles, I will show
how the source decreases it power output to compensate for destructive
interference and increases it power output to compensate for
constructive interference.

Unfortunately, the circuit is intended to illustrate the absence of

[AVERAGE] interference under special circumstances but an instant analysis shows

that all the power can not be accounted for.

Not surprising since there is no conservation of power principle.

We can only conclude that

[instantaneous] interference is present. Not good because the circuit was intended to

illustrate a case of NO [AVERAGE] interference.

I took the liberty of adding adjectives in brackets[*] to your
above statements. It doesn't matter about the instantaneous values
of power since not only do they not have to be conserved, but they
are also "of limited usefulness", according to Eugene Hecht, since
the actual energy content of instantaneous power is undefined even
when the instantaneous power is defined.

The circuit is very useful to investigate interference more carefully because on the AVERAGE,

the interference IS zero. Using spreadsheets, we can see how the
interference both adds and

subtracts from the instantaneous applied voltage, resulting in cycling
variations in the power

applied to the resistor and other circuit elements. A very instructive
exercise.

Instructive as long as we remember that a conservation of power
principle doesn't exist and therefore, equations based on instantaneous
powers do not have to balance. The joules, not the watts, are what must
balance.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Cecil Moore wrote:
Roger Sparks wrote:
You write "The only other device in the entire system capable
of dissipation is the source resistor." which is a correct statement.

Therefore, all power dissipated in the circuit must be dissipated
in the load resistor and the source resistor because there is
nowhere else for it to go. Since the reflected power is not
dissipated in the load, by definition, it has to be dissipated
in the source resistor but not at the exact time of its arrival.
There is nothing wrong with delaying power dissipation for 90
degrees of the cycle. In Parts 2 and 3 of my articles, I will show
how the source decreases it power output to compensate for destructive
interference and increases it power output to compensate for
constructive interference.

Unfortunately, the circuit is intended to illustrate the absence of

[AVERAGE] interference under special circumstances but an instant
analysis shows

that all the power can not be accounted for.

Not surprising since there is no conservation of power principle.

The concept of a wave is energy located at a predicted place after some
time period. That is a concept of conservation of power.

We can only conclude that

[instantaneous] interference is present. Not good because the circuit
was intended to

illustrate a case of NO [AVERAGE] interference.

I took the liberty of adding adjectives in brackets[*] to your
above statements. It doesn't matter about the instantaneous values
of power since not only do they not have to be conserved, but they
are also "of limited usefulness", according to Eugene Hecht, since
the actual energy content of instantaneous power is undefined even
when the instantaneous power is defined.

The circuit is very useful to investigate interference more carefully
because on the AVERAGE,

the interference IS zero. Using spreadsheets, we can see how the
interference both adds and

subtracts from the instantaneous applied voltage, resulting in cycling
variations in the power

applied to the resistor and other circuit elements. A very instructive
exercise.

Instructive as long as we remember that a conservation of power
principle doesn't exist and therefore, equations based on instantaneous
powers do not have to balance. The joules, not the watts, are what must
balance.

Forget the conservation of power at your own peril, because we need to
depend upon the predictability of waves of energy acting over time to
solve these problems. When the instantaneous powers do not balance, we
know that we do not yet have the complete solution or complete circuit.


73, Roger, W7WKB